Implicit differentiation is just like regular differentiation. We keep respect to x.
For x²-y²=16
First we do differentiation by parts. What is the derivative of x²? 2x
How about y² though? This will make use of chain rule. First we have power rule, so 2y.
BUT remember this is y as a function of x. So whenever we derive y, we have to multiply it by dy/dx
Then the derivative of 16 is just 0.
So we will end up with
2x-2y*dy/dx = 0
-2y*dy/dx=-2x
dy/dx = 2x/2y

x^2-y^2=16; y'=?
as previously stated the derivative of a constant is zero. so, the right side of the equation turns into zero once the derivative of 16 is obtained.
on the other hand,
x^2-y^2 makes use of the power rule,
the power rule is
nx^(n-1)
so applying to the x^2, it becomes
2x
the purpose of implicit differentiation is obtaining the derivative without foreknowledge of the value of y. the problem you're solving may not be the best example to emphasize the beauty of this method, but we nevertheless can implement it. therefore, we will do this
y^2 also makes use of the power rule and becomes
2y and since we do not know what y is, we will make it
2y * y' (other notations may be used such as the dy/dx, but I just prefer y' for simplicity); this just simply means the derivative of y^2 multiplied by the unknown derivative of y, whatever value it may be.
now, let us all put it together
2x - 2y * y' = 0
we want the implicit differentation y' this is the easy part - the algebra
so it then becomes
y' = 2x/2y
you can then plug your coordinate (5, 3) into their respective places.
y' = 2(5) / 2 (3)
y' = 10 / 6
y' = 5/3
looking at the graph,
http://www.wolframalpha.com/input/?i=x%5E2-y%5E2%3D16
we are confident that the slope of our tangent line at (5, 3) should be positive, and that is exactly we have.
I am not sure if this particular example has been emphasized to you.
derivative of x^2 is 2x * x' and since the derivative of x is 1, it is usually left out since 2x * 1 is 2x
this is the basis of implicit differentiation. we do not know the actual value of what we are differentiating.